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[tex]\frac{{\partial}^{2}f}{{\partial x}^{2}}+\frac{{\partial}^{2}f}{{\partial y}^{2}}=0[/tex]

Thanks in advance.

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[tex]\frac{{\partial}^{2}f}{{\partial x}^{2}}+\frac{{\partial}^{2}f}{{\partial y}^{2}}=0[/tex]

Thanks in advance.

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You really have to give the boundary conditions when dealing with Laplace equation, and in general for PDEs, boundary and initial conditions.

The general solution is:

[tex]f(x,y)=g(z)+h(\overline{z}),\quad z=x+iy[/tex]

where g and h are arbitrary [itex]C^2[/itex] functions of a single variable z=x+iy. We can show this by factoring the equation:

[tex]\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right) f=\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)=\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)u=0[/tex]

where:

[tex]u=\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}[/tex]

So first solve:

[tex]u_t-iu_x=0[/tex]

which is an easy first-order equation which turns out to be u=g(x+it) then substitute that into:

[tex]f_x+if_y=g(x+it)[/tex]

another easy one to obtain so that the general solution is:

[tex]f(x,y)=g(x+it)+h(x-it)[/tex]

Note that means the general solution is not analytic since it contains [itex]\overline{z}[/itex] but analyticity is not a requirement for the solution but only differentiability.

[tex]f(x,y)=g(z)+h(\overline{z}),\quad z=x+iy[/tex]

where g and h are arbitrary [itex]C^2[/itex] functions of a single variable z=x+iy. We can show this by factoring the equation:

[tex]\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right) f=\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)=\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)u=0[/tex]

where:

[tex]u=\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}[/tex]

So first solve:

[tex]u_t-iu_x=0[/tex]

which is an easy first-order equation which turns out to be u=g(x+it) then substitute that into:

[tex]f_x+if_y=g(x+it)[/tex]

another easy one to obtain so that the general solution is:

[tex]f(x,y)=g(x+it)+h(x-it)[/tex]

Note that means the general solution is not analytic since it contains [itex]\overline{z}[/itex] but analyticity is not a requirement for the solution but only differentiability.

Last edited:

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Thanks.The general solution is:

[tex]f(x,y)=g(z)+h(\overline{z}),\quad z=x+iy[/tex]

where g and h are arbitrary [itex]C^2[/itex] functions of a single variable z=x+iy. We can show this by factoring the equation:

[tex]\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right) f=\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)=\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)u=0[/tex]

where:

[tex]u=\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}[/tex]

So first solve:

[tex]u_t-iu_x=0[/tex]

which is an easy first-order equation which turns out to be u=g(x+it) then substitute that into:

[tex]f_x+if_y=g(x+it)[/tex]

another easy one to obtain so that the general solution is:

[tex]f(x,y)=g(x+it)+h(x-it)[/tex]

Note that means the general solution is not analytic since it contains [itex]\overline{z}[/itex] but analyticity is not a requirement for the solution but only differentiability.

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